Sums of coefficients

Put $x = 1$: the sum of all the coefficients

Write the expansion again:

Put $x = 1$. The left side is $2^n$, and the right side is the sum of the coefficients:

Row $8$ adds up to $1 + 8 + 28 + 56 + 70 + 56 + 28 + 8 + 1 = 256 = 2^8$.

The same trick works for any polynomial. To add up all its coefficients, put $x = 1$.

Example 1. The coefficients of $(2x + 3)^5$ add up to $(2 + 3)^5 = 3125$.

Example 2. In $(x^2 + 3x - 1)^6$, putting $x = 1$ gives $(1 + 3 - 1)^6 = 3^6 = 729$. We never expanded the bracket.

Put $x = -1$: the alternating sum

Now put $x = -1$ in $(1 + x)^n$. The left side is $0$ when $n \ge 1$. The right side is

So the coefficients with a plus sign add up to the same as those with a minus sign. The plus signs belong to the odd-placed terms, the 1st, 3rd, 5th and so on. The minus signs belong to the even-placed terms, the 2nd, 4th, 6th and so on.

Together they make $2^n$, and they are equal. So each is half:

For row $8$, $1 + 28 + 70 + 28 + 1 = 128$ and $8 + 56 + 56 + 8 = 128$.

Choose any row. The two sums are always equal, and the sum of all the coefficients is a power of $2$.

The same idea for a polynomial

Let $f(x) = a_0 + a_1 x + a_2 x^2 + \cdots$ be any polynomial. Then $f(1)$ adds all the coefficients, and $f(-1)$ adds the even powers and subtracts the odd ones. So

Example. Take $f(x) = (x + 2)^5$. Then $f(1) = 3^5 = 243$ and $f(-1) = 1^5 = 1$. The coefficients of the even powers of $x$ add up to $\dfrac{243 + 1}{2} = 122$, and those of the odd powers to $\dfrac{243 - 1}{2} = 121$.

Check. Expanding gives $32 + 80x + 80x^2 + 40x^3 + 10x^4 + x^5$. The even powers give $32 + 80 + 10 = 122$, and the odd powers give $80 + 40 + 1 = 121$.

A sum with $r \cdot {}^{n}C_{r}$

Find ${}^{n}C_{1} + 2\,{}^{n}C_{2} + 3\,{}^{n}C_{3} + \cdots + n\,{}^{n}C_{n}$.

Step 1. Use the identity. $r \cdot {}^{n}C_{r} = n \cdot {}^{n - 1}C_{r - 1}$. So the sum is

Step 2. Add the brackets. The coefficients in the bracket belong to row $n - 1$, so they add up to $2^{n - 1}$.

Answer.

Check for $n = 6$. The terms are $6, 30, 60, 60, 30, 6$, which add up to $192$. The formula gives $6 \times 2^5 = 192$.

A sum with a division

Find ${}^{n}C_{0} + \dfrac{{}^{n}C_{1}}{2} + \dfrac{{}^{n}C_{2}}{3} + \cdots + \dfrac{{}^{n}C_{n}}{n + 1}$.

Step 1. A new identity. ${}^{n + 1}C_{r + 1} = \dfrac{n + 1}{r + 1} \cdot {}^{n}C_{r}$. So $\dfrac{{}^{n}C_{r}}{r + 1} = \dfrac{{}^{n + 1}C_{r + 1}}{n + 1}$.

Step 2. Add. The numerators run through row $n + 1$, except its first entry:

Check for $n = 5$. The terms are $1, 2.5, 3.33\ldots, 2.5, 1, 0.1666\ldots$, which add up to $10.5$. The formula gives $\dfrac{64 - 1}{6} = 10.5$.

The sum of the squares

The squares of one row add up to a coefficient in a longer row:

The reason is that $(1 + x)^n (1 + x)^n = (1 + x)^{2n}$. The coefficient of $x^n$ on the right is ${}^{2n}C_{n}$. On the left it is the sum of products ${}^{n}C_{r} \cdot {}^{n}C_{n - r}$, and ${}^{n}C_{n - r} = {}^{n}C_{r}$ by symmetry.

For $n = 4$: $1 + 16 + 36 + 16 + 1 = 70 = {}^{8}C_{4}$.

Practise with new numbers

Try it in Python

The program prints, for every row from $1$ to $10$, the sum of all the coefficients, the two half-sums, their difference, and the sum of $r \cdot {}^{n}C_{r}$ beside $n \cdot 2^{n - 1}$.

Common mistakes

• Expanding first. To add the coefficients of a polynomial, put $x = 1$. There is no need to expand.
• Forgetting that $x = -1$ needs $n \ge 1$. For $n = 0$ the alternating sum is $1$, not $0$.
• Calling the half-sum $2^n$. Each half is $2^{n - 1}$.
• Mixing up the two half-sums. The 1st, 3rd and 5th terms hold ${}^{n}C_{0}$, ${}^{n}C_{2}$ and ${}^{n}C_{4}$, the even values of $r$.
• Using the identity with the wrong row. After $r \cdot {}^{n}C_{r} = n \cdot {}^{n - 1}C_{r - 1}$, the new coefficients are in row $n - 1$, so they add up to $2^{n - 1}$.

Remember these

• The sum of the coefficients of $(1 + x)^n$ is $2^n$.
• The sum with alternating signs is $0$, so the odd-placed and the even-placed terms each add up to $2^{n - 1}$.
• For a polynomial $f$, put $x = 1$ for the total and $x = -1$ for the alternating sum.
• $\displaystyle\sum_{r = 0}^{n} r \cdot {}^{n}C_{r} = n \cdot 2^{n - 1}$.
• $\displaystyle\sum_{r = 0}^{n} \dfrac{{}^{n}C_{r}}{r + 1} = \dfrac{2^{n + 1} - 1}{n + 1}$.
• $\displaystyle\sum_{r = 0}^{n} \left({}^{n}C_{r}\right)^2 = {}^{2n}C_{n}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the sum of the coefficients in the expansion of $(1 + x)^7$.

2. Find the sum of the coefficients in the expansion of $(3x - 1)^4$.

3. Find the sum of the odd-placed coefficients in the expansion of $(1 + x)^9$.

4. Find the value of $\displaystyle\sum_{r = 0}^{5} r \cdot {}^{5}C_{r}$.

5. Find the value of $\left({}^{3}C_{0}\right)^2 + \left({}^{3}C_{1}\right)^2 + \left({}^{3}C_{2}\right)^2 + \left({}^{3}C_{3}\right)^2$.